Lemma 55.14.2. Let $R$ be a discrete valuation ring. Let $X$ be a scheme which is at-worst-nodal of relative dimension $1$ over $R$. Let $x \in X$ be a point of the special fibre of $X$ over $R$. Then there exists a commutative diagram

$\xymatrix{ X \ar[d] & U \ar[r] \ar[d] \ar[l] & \mathop{\mathrm{Spec}}(A) \ar[dl] \\ \mathop{\mathrm{Spec}}(R) & \mathop{\mathrm{Spec}}(R') \ar[l] }$

where $R \subset R'$ is an étale extension of discrete valuation rings, the morphism $U \to X$ is étale, the morphism $U \to \mathop{\mathrm{Spec}}(A)$ is étale, there is a point $x' \in U$ mapping to $x$, and

$A = R'[u, v]/(uv) \quad \text{or}\quad A = R'[u, v]/(uv - \pi ^ n)$

where $n \geq 0$ and $\pi \in R'$ is a uniformizer.

Proof. We have already proved this lemma in much greater generality, see Algebraic Curves, Lemma 53.20.12. All we have to do here is to translate the statement given there into the statement given above.

First, if the morphism $X \to \mathop{\mathrm{Spec}}(R)$ is smooth at $x$, then we can find an étale morphism $U \to \mathbf{A}^1_ R = \mathop{\mathrm{Spec}}(R[u])$ for some affine open neighbourhood $U \subset X$ of $x$. This is Morphisms, Lemma 29.36.20. After replacing the coordinate $u$ by $u + 1$ if necessary, we may assume that $x$ maps to a point in the standard open $D(u) \subset \mathbf{A}^1_ R$. Then $D(u) = \mathop{\mathrm{Spec}}(A)$ with $A = R[u, v]/(uv - 1)$ and we see that the result is true in this case.

Next, assume that $x$ is a singular point of the fibre. Then we may apply Algebraic Curves, Lemma 53.20.12 to get a diagram

$\xymatrix{ X \ar[d] & U \ar[rr] \ar[l] \ar[rd] & & W \ar[r] \ar[ld] & \mathop{\mathrm{Spec}}(\mathbf{Z}[u, v, a]/(uv - a)) \ar[d] \\ \mathop{\mathrm{Spec}}(R) & & V \ar[ll] \ar[rr] & & \mathop{\mathrm{Spec}}(\mathbf{Z}[a]) }$

with all the properties mentioned in the statement of the cited lemma. Let $x' \in U$ be the point mapping to $x$ promised by the lemma. First we shrink $V$ to an affine neighbourhood of the image of $x'$. Say $V = \mathop{\mathrm{Spec}}(R')$. Then $R \to R'$ is étale. Since $R$ is a discrete valuation ring, we see that $R'$ is a finite product of quasi-local Dedekind domains (use More on Algebra, Lemma 15.44.4). Hence (for example using prime avoidance) we find a standard open $D(f) \subset V = \mathop{\mathrm{Spec}}(R')$ containing the image of $x'$ such that $R'_ f$ is a discrete valuation ring. Replacing $R'$ by $R'_ f$ we reach the situation where $V = \mathop{\mathrm{Spec}}(R')$ with $R \subset R'$ an étale extension of discrete valuation rings (extensions of discrete valuation rings are defined in More on Algebra, Definition 15.111.1).

The morphism $V \to \mathop{\mathrm{Spec}}(\mathbf{Z}[a])$ is determined by the image $h$ of $a$ in $R'$. Then $W = \mathop{\mathrm{Spec}}(R'[u, v]/(uv - h))$. Thus the lemma holds with $A = R'[u, v]/(uv - h)$. If $h = 0$ then we clearly obtain the first case mentioned in the lemma. If $h \not= 0$ then we may write $h = \epsilon \pi ^ n$ for some $n \geq 0$ where $\epsilon$ is a unit of $R'$. Changing coordinates $u_{new} = \epsilon u$ and $v_{new} = v$ we obtain the second isomorphism type of $A$ listed in the lemma. $\square$

Comment #5403 by Laurent Moret-Bailly on

The statement says nothing about the morphism $U\to\mathrm{Spec}(A)$.

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